Part one
D=⎝⎛121111213⎠⎞
Characteristic equation det∣∣D−λI∣∣=0
Det⎝⎛1−λ2111−λ1213−λ⎠⎞
Characteristic polynomial is of the form;
λ3−A1λ2+A2λ−A3=0
Where A1= sum of main diagonal element
=1+1+3=5
A2= sum of minors of the main diagonal element
A2=∣∣1113∣∣+∣∣1123∣∣+∣∣1211∣∣=2+1−1=2
A3=det(D)=1∣∣1113∣∣−1∣∣2113∣∣+2∣∣2111∣∣
=1(2)−1(5)+2(1)
=−1
Characteristic polynomial is
λ3−5λ2+2λ+1=0
Let λ3−5λ2+2λ+1=0
f(λ)=λ3−5λ2+2λ+1
1st iteration
λ0=5 and λ1=4
f(λ0)=f(s)=11
f(λ1)=f(4)=−7
∴λ2=5−11 −7−114−5=4.3889
f(λ2)=−1.9937
2nd iteration
λ1=4 and λ2=4.3889
f(λ1)=f(4)=−7
f(λ2)=f(4.3889)=−1.9937
λ3=4−(−7) −1.9937−(−7)4.3889−4=4.5438
3rd iteration
λ2=4.3889 and λ3=4.5438
f(λ2)=−1.9937
f(λ3)=f(4.5438)=0.6688
λ4=4.3889−(−1.9937)
0.6688−−1.99374.5438−4.3889
=4.5049
f(λ4)=f(4.5049)=−0.0378
4th iteration
λ3=4.5438 and λ4=4.5049
f(λ3)=0.6688
f(λ4)=−0.0378
λ5=4.5438−0.6688
−0.0378−0.66884.5049−4.5438
=4.5070
f(λ5)=−0.0003
Approximate root =4.507
Part two
Q -R method
Let a1=[121] a2=[111]
and
a3=[213]
Let orthogonal set =(b1b2b3)
Let orthonormal set =(q1q2q3)
Let b1=[121],
∣∣b1∣∣=12+22+12=2.45
q1=2.451[121]=[0.41,0.82,0.41]
b2=[111]−
<(1,1,1),(0.41,0.82,0.41)>(0.41,0.82,0.41)
=(0.33,−0.33,0.33)
∣∣b2∣∣=0.332+0.332+0.332=
0.58
q2=0.581[0.33,−0.330.33]=
(0.58,−0.58,0.58)
b3=(2,1,3) −
−<(2,1,3),(0.41,0.82,0.41)>
(0.41,0.82,0.41)
−<(2,1,3),(0.58,−0.58,0.58)>
b3=(−0.5,0,0.5)∣∣b3∣∣=
0.52+02+0.52=0.71
q3=0.711(−0.5,0,0.5)=
(−0.71,0,0.71)
∴Q=[q1q2q3]=
⎣⎡0.410.820.410.58−0.580.58−0.7100.71⎦⎤
R=QTA=⎝⎛0.410.58−0.710.82−0.5800.410.580.71⎠⎞⎝⎛121111213⎠⎞
=⎝⎛2.45001.630.5802.862.310.71⎠⎞
Denoting the above R as R0 and the Q as Q0
Let A1=R0Q0
Let A1=Q1R1 be QR factorization of A1 and similarly create A2=R1Q1
This process is continued until Am is created such that Am=QmRm and Am1=RmQm are equal.
At convergence, the diagonal entries below the main diagonal are sufficiently small.
The diagonal entries of Am will be eigenvalues of matrix A
The expected values will be approximate to 0.28514,0.77812and4.50701
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